Bound $|a-b|^p $ by $|a^p - b^p| $.

Question

I assume that $a>0$ and $b>0.$

I want to know if there is a bound of the type

$|a-b|^p \leq K |a^p - b^p|$

where $p>1$ and $K$ is a constant depending on $p.$ I would like to know the explicit form of the constant $K$ if there exists one. I know that

$(a+b)^p \leq 2^p (a^p + b^p)$

but this doesn't help.

Answer 1

Idea: Since $b\ne 0$ you can make $x=a/b$, so we get $$|x-1|^p\leq K|x^p-1|$$

Let $x\ne 1$ and make $$f(x)= {|x-1|^p\over |x^p-1|} = {(x-1)^p\over x^p-1}$$

Now calculate maximum of $f$. Since $$f'(x)= {p(x^{p-1}-1)(x-1)^{p-1}\over (x^p-1)^2}$$

We can finish here fast if $p$ is integer.

If $p$ is odd then for $x\in (0,1)$ $f$ is decreasing and for $x>1$ $f$ is increasing. Since $f(0) = 1$ and $\lim_{x\to \infty}=1$ we have $f_{max} \leq 1$.

If $p$ is even then $f'>0$ so $f$ is increasing. Since $\lim_{x\to \infty}=1$ we have $f_{max} \leq 1$.

So $K=1$ for $p\in \mathbb{N}$.

Answer 2

Hint: Fix $a>0.$ On $[a,\infty)$ consider the functions

$$f(x) = (x-a)^p,\,\, g(x) = x^p-a^p.$$

We have $f(a)=0=g(a).$ Show $f'\le g'$ on $[a,\infty).$ (I'm getting $K=1$ works for all $p,$ and that it's the smallest such constant.)